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arXiv 1807.01064cs.ITcs.CCmath.COmath.IT
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最优局部可修码能有多长?

How long can optimal locally repairable codes be?

Venkatesan Guruswami, Chaoping Xing, Chen Yuan

摘要:

具有局部性$r$的局部可修复码(LRC)允许仅使用$r$其它码字符号来恢复任何被擦除的码字符号。一个单重子类型的界限决定了LRC的尺寸和距离之间的最佳可能的折衷-达到这种折衷的LRC被认为是\emph{optimal}。这种最佳的LRC已经被构造成在块长度上线性增长的字母表。然而,与经典的Singleton界不同,我们不知道字母表大小的线性增长是否是必要的,或者即使字母表需要随着块长度而增长。事实上,对于小的码距$3,4$,任意长的最优LRC在固定字母表上是已知的。在这里,我们证明了距离$d \ge 5$,代码长度$n$的最佳LRC在一个字母表的大小$q$必须是最多大约$O(d q^3)$。对于$d=5$情形,我们的上界是$O(q^2)$。我们补充这些界限显示存在的最佳LRC长度$Ω_{d,r}(q^{1+1/\lfloor(d-3)/2\rfloor})$时,$d \le r+2$。当$d=5$时,这些边界匹配,从而将$n=Θ(q^2)$固定为这种情况下最佳LRC的渐近最大长度。

英文摘要:

A locally repairable code (LRC) with locality $r$ allows for the recovery of any erased codeword symbol using only $r$ other codeword symbols. A Singleton-type bound dictates the best possible trade-off between the dimension and distance of LRCs --- an LRC attaining this trade-off is deemed \emph{optimal}. Such optimal LRCs have been constructed over alphabets growing linearly in the block length. Unlike the classical Singleton bound, however, it was not known if such a linear growth in the alphabet size is necessary, or for that matter even if the alphabet needs to grow at all with the block length. Indeed, for small code distances $3,4$, arbitrarily long optimal LRCs were known over fixed alphabets. Here, we prove that for distances $d \ge 5$, the code length $n$ of an optimal LRC over an alphabet of size $q$ must be at most roughly $O(d q^3)$. For the case $d=5$, our upper bound is $O(q^2)$. We complement these bounds by showing the existence of optimal LRCs of length $Ω_{d,r}(q^{1+1/\lfloor(d-3)/2\rfloor})$ when $d \le r+2$. These bounds match when $d=5$, thus pinning down $n=Θ(q^2)$ as the asymptotically largest length of an optimal LRC for this case.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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